Answer
See below
Work Step by Step
Obtain $T(x^2+x)=2x+1\\\
T(1)=0\\
T(x)=1$
then we have:
$T(x^2+x)=2x+1=-1.1+2.(1+x)\\ T(1)=0=0.1+0.(1+x)\\ T(x)=1=1.1+0.(1+x)$
Thus,
$T[(x^2+x)]_C=\begin{bmatrix}
=1\\
2 \end{bmatrix}\\
T[(1)]_C=\begin{bmatrix}
0\\
0 \end{bmatrix}\\
T[(x)]_C=\begin{bmatrix}
1\\
0 \end{bmatrix}$
Hence, $[T]_B^C=\begin{bmatrix}
-1 & 0 & 1 \\
2 & 0 & 0 \end{bmatrix}$
We can notice that $[v]_B=\begin{bmatrix}
1 & -3 & -1 \end{bmatrix}$
From Theorem 6.4.5,
$$[T(v)]_C=[T]^C_B[v]_B\\
\rightarrow [T(v)]_C=\begin{bmatrix}
-1 & 0 & 1 \\
2 & 0 & 0 \end{bmatrix}\begin{bmatrix}
1 & -3 & -1\end{bmatrix}=\begin{bmatrix}
-2\\ -2\end{bmatrix}\\
\rightarrow T(v)=-2.1+2(1+x)=2x$$
